Use differentials to approximate `sqrt(25. 2)`

Explanation

`y= f(x) = sqrtx`

`/_ y = f(x +/_ x) - f(x)`

`x= 25 & /_ x= 0.2`

`=> f(x + /_ x) = f(x) + /_ y`

`f(x) = f(25) = 5`

`/_y = dy = (dy/dx)/_x = (d(sqrtx))/(dx) xx 0.2 `

`/_ y = 1/(2 sqrtx) xx 0.2`

`= 0.1/5 = 1/50`

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